Class 9 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 10 - Quadrilaterals
Quadrilaterals Exercise MCQ
Solution 1
Solution 2
Solution 3
Solution 4
Correct option: (a)
ABCD is a rhombus.
⇒ AD ∥ BC and AC is the transversal.
⇒ ∠DAC = ∠ACB (alternate angles)
⇒ ∠DAC = 50°
In ΔAOD, by angle sum property,
∠AOD + ∠DAO + ∠ADO = 180°
⇒ 90° + ∠50° + ∠ADO = 180°
⇒ ∠ADO = 40°
⇒ ∠ADB = 40°
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Correct option: (b)
∠DAO + ∠OAB = ∠DAB
⇒ ∠DAO + 35° = 90°
⇒ ∠DAO = 55°
ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.
OA = OD
⇒ ∠ODA = ∠DAO (angles opposte to equal sides are equal)
⇒ ∠ODA = 55°
In DODA, by angle sum property,
∠ODA + ∠DAO + ∠AOD = 180°
⇒ 55° + ∠55° + ∠AOD = 180°
⇒ ∠AOD = 70°
Solution 10
Solution 11
Solution 12
Solution 13
Correct option: (c)
The bisectors of the angles of a parallelogram enclose a rectangle.
Solution 14
Correct option: (d)
In ΔAPB, by angle sum property,
∠APB + ∠PAB + ∠PBA = 180°
In ΔCRD, by angle sum property,
∠CRD + ∠RDC + ∠RCD = 180°
Now, ∠SPQ + ∠SRQ = ∠APB + ∠CRD
= 360° - 180°
= 180°
Now, ∠PSR + ∠PQR = 360° - (∠SPQ + ∠SRQ)
= 360° - 180°
= 180°
Hence, PQRS is a quadrilateral whose opposite angles are supplementary.
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Correct option: (d)
In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
In ΔADC, R and S are the mid-points of sides CD and AD respectively.
From (i) and (ii),
PQ ∥ RS and PQ = RS
Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.
So, PQRS is a parallelogram.
Let the diagonals AC and BD intersect at O.
Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.
Thus, in quadrilateral PMON, PM ∥ NO and PN ∥ MO.
⇒ PMON is a parallelogram.
⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)
⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)
⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)
⇒ ∠QPS = 90°
Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.
Hence, PQRS is a rectangle if AC ⊥ BD.
Solution 21
Correct option: (c)
In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.
In ΔADC, S and R are the mid-points of sides AD and CD respectively.
In ΔABD, P and S are the mid-points of sides AB and AD respectively.
⇒ PQ ∥ RS and QR ∥ SP [From (i), (ii), (iii) and (iv)]
Thus, PQRS is a parallelogram.
Now, AC = BD (given)
⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]
Hence, PQRS is a rhombus if diagonals of ABCD are equal.
Solution 22
Correct option: (d)
In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.
In ΔADC, S and R are the mid-points of sides AD and CD respectively.
In ΔABD, P and S are the mid-points of sides AB and AD respectively.
⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]
Thus, PQRS is a parallelogram.
Now, AC = BD (given)
⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]
Let the diagonals AC and BD intersect at O.
Now,
Thus, in quadrilateral PMON, PM || NO and PN || MO.
⇒ PMON is a parallelogram.
⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)
⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)
⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)
⇒ ∠QPS = 90°
Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.
Hence, PQRS is a square if diagonals of ABCD are equal and perpendicular.
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Correct option: (b)
In ΔABC, D and E are the mid-points of sides AB and AC respectively.
Hence, DENM is a parallelogram.
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39
Solution 40
Solution 41
Solution 42
Since opposite angles of quadrilateral are equal, PQRS is a parallelogram.
⇒ PQ = SR (opposite sides of parallelogram are equal)
⇒ PQ = 2 cm
Solution 43
The given statement is false.
Diagonals of a parallelogram bisect each other.
Solution 44
In quadrilateral PQRS, ∠P and ∠S are adjacent angles.
Since the sum of adjacent angles ≠ 180°, PQRS is not a parallelogram.
Hence, PQRS is a trapezium.
Solution 45
The given statement is false.
We know that the sum of all the four angles of a quadrilateral is 360°.
If all the angles of a quadrilateral are acute, the sum will be less than 360°.
Solution 46
The given statement is true.
We know that the sum of all the four angles of a quadrilateral is 360°.
If all the angles of a quadrilateral are right angles,
Sum of all angles of a quadrilateral = 4 × 90° = 360°
Solution 47
The given statement is false.
We know that the sum of all the four angles of a quadrilateral is 360°.
If all the angles of a quadrilateral are obtuse, the sum will be more than 360°.
Solution 48
We know that the sum of all the four angles of a quadrilateral is 360°.
Here,
70° + 115° + 60° + 120° = 365° ≠ 360°
Hence, we cannot form a quadrilateral with given angles.
Solution 49
A quadrilateral whose all angles are equal is a rectangle.
Solution 50
D and E are respectively the midpoints of the sides AB and BC of ΔABC.
Thus, by mid-point theorem, we have
Solution 51
Since the diagonals PR and QS of quadrilateral PQRS bisect each, PQRS is a parallelogram.
Now, adjacent angles of parallelogram are supplementary.
⇒ ∠Q + ∠R = 180°
⇒ 56° + ∠R = 180°
⇒ ∠R = 124°
Solution 52
AFDE is a parallelogram
⇒ AF = ED …(i)
BDEF is a parallelogram.
⇒ FB = ED …(ii)
From (i) and (ii),
AF = FB
Solution 53
Correct option: (a)
If the diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a parallelogram.
So, I gives the answer.
If the diagonals are equal, then the quad. ABCD is a parallelogram.
So, II gives the answer.
Solution 54
Correct option: (c)
If the quad. ABCD is a ‖gm, it could be a rectangle or square or rhombus.
So, statement I is not sufficient to answer the question.
If the diagonals AC and BD are perpendicular to each other, then the ‖gm could be a square or rhombus.
So, statement II is not sufficient to answer the question.
However, if the statements are combined, then the quad. ABCD is a rhombus.
Solution 55
Correct option: (c)
If the diagonals of a ‖gm ABCD are equal, then ‖gm ABCD could either be a rectangle or a square.
If the diagonals of the ‖gm ABCD intersect at right angles, then the ‖gm ABCD could be a square or a rhombus.
However, if both the statements are combined, then ‖gm ABCD will be a square.
Solution 56
Correct option: (b)
If the opposite sides of a quad. ABCD are equal, the quadrilateral is a parallelogram.
If the opposite angles are equal, then the quad. ABCD is a parallelogram.
Solution 57
Solution 58
The Reason (R) is true and is the correct explanation for the Assertion (A).
Solution 59
Solution 60
Solution 61
Solution 62
Solution 63
Quadrilaterals Exercise Ex. 10B
Solution 1
Solution 2
Solution 3
ABCD is a parallelogram.
Hence, AD || BC.
⇒ ∠DAM = ∠AMB (alternate angles)
⇒ ∠BAM = ∠AMB (since ∠BAM = ∠DAM)
⇒ BM = AB (sides opposite to equal angles are equal)
But, AB = CD (opposite sides of a parallelogram)
⇒ BM = AB = CD ….(i)
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Let the altitude from D to the side AB bisect AB at point P.
Join BD.
In ΔAMD and ΔBMD,
AM = BM (M is the mid-point of AB)
∠AMD = ∠BMD (Each 90°)
MD = MD (common)
∴ ΔAMD ≅ ΔBMD (by SAS congruence criterion)
⇒ AD = BD (c.p.c.t.)
But, AD = AB (sides of a rhombus)
⇒ AD = AB = BD
⇒ ΔADB is an equilateral triangle.
⇒ ∠A = 60°
⇒ ∠C = ∠A = 60° (opposite angles are equal)
⇒ ∠B = 180° - ∠A = 180° - 60° = 120°
⇒ ∠D = ∠B = 120°
Hence, in rhombus ABCD, ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°.
Solution 15
Solution 16
In ΔABC and ΔADC,
AB = AD (sides of a rhombus are equal)
BC = CD (sides of a rhombus are equal)
AC = AC (common)
∴ ΔABC ≅ ΔADC (by SSS congruence criterion)
⇒ ∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)
⇒ AC bisects ∠A as well as ∠C.
Similarly,
In ΔBAD and ΔBCD,
AB = BC (sides of a rhombus are equal)
AD = CD (sides of a rhombus are equal)
BD = BD (common)
∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)
⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)
⇒ BD bisects ∠B as well as ∠D.
Solution 17
In ΔAMO and ΔCNO
∠MAO = ∠NCO (AB ∥ CD, alternate angles)
AM = CN (given)
∠AOM = ∠CON (vertically opposite angles)
∴ ΔAMO ≅ ΔCNO (by ASA congruence criterion)
⇒ AO = CO and MO = NO (c.p.c.t.)
⇒ AC and MN bisect each other.
Solution 18
Solution 19
Solution 20
∠DCM = ∠DCN + ∠MCN
⇒ 90° = ∠DCN + 60°
⇒ ∠DCN = 30°
In ΔDCN,
∠DNC + ∠DCN + ∠D = 180°
⇒ 90° + 30° + ∠D = 180°
⇒ ∠D = 60°
⇒ ∠B = ∠D = 60° (opposite angles of parallelogram are equal)
⇒ ∠A = 180° - ∠B = 180° - 60° = 120°
⇒ ∠C = ∠A = 120°
Thus, the angles of a parallelogram are 60°, 120°, 60° and 120°.
Solution 21
(i) ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.
⇒ ∠BAC = ∠DAC ….(i)
And ∠BCA = ∠DCA ….(ii)
Since every rectangle is a parallelogram, therefore
AB ∥ DC and AC is the transversal.
⇒ ∠BAC = ∠DCA (alternate angles)
⇒ ∠DAC = ∠DCA [From (i)]
Thus, in ΔADC,
AD = CD (opposite sides of equal angles are equal)
But, AD = BC and CD = AB (ABCD is a rectangle)
⇒ AB = BC = CD = AD
Hence, ABCD is a square.
(ii) In ΔBAD and ΔBCD,
AB = CD
AD = BC
BD = BD
∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)
⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)
Hence, diagonal BD bisects ∠B as well as ∠D.
Solution 22
Solution 23
Solution 24
l ∥ m and t is a transversal.
⇒ ∠APR = ∠PRD (alternate angles)
⇒ ∠SPR = ∠PRQ (PS and RQ are the bisectors of ∠APR and ∠PRD)
Thus, PR intersects PS and RQ at P and R respectively such that ∠SPR = ∠PRQ i.e., alternate angles are equal.
⇒ PS ∥ RQ
Similarly, we have SR ∥ PQ.
Hence, PQRS is a parallelogram.
Now, ∠BPR + ∠PRD = 180° (interior angles are supplementary)
⇒ 2∠QPR + 2∠QRP = 180° (PQ and RQ are the bisectors of ∠BPR and ∠PRD)
⇒ ∠QPR + ∠QRP = 90°
In ΔPQR, by angle sum property,
∠PQR + ∠QPR + ∠QRP = 180°
⇒ ∠PQR + 90° = 180°
⇒ ∠PQR = 90°
Since PQRS is a parallelogram,
∠PQR = ∠PSR
⇒ ∠PSR = 90°
Now, ∠SPQ + ∠PQR = 180° (adjacent angles in a parallelogram are supplementary)
⇒ ∠SPQ + 90° = 180°
⇒ ∠SPQ = 90°
⇒ ∠SRQ = 90°
Thus, all the interior angles of quadrilateral PQRS are right angles.
Hence, PQRS is a rectangle.
Solution 25
AK = BL = CM = DN (given)
⇒ BK = CL = DM = AN (i)(since ABCD is a square)
In ΔAKN and ΔBLK,
AK = BL (given)
∠A = ∠B (Each 90°)
AN = BK [From (i)]
∴ ΔAKN ≅ ΔBLK (by SAS congruence criterion)
⇒ ∠AKN = ∠BLK and ∠ANK = ∠BKL (c.p.c.t.)
But, ∠AKN + ∠ANK = 90° and ∠BLK + ∠BKL = 90°
⇒ ∠AKN + ∠ANK + ∠BLK + ∠BKL = 90° + 90°
⇒ 2∠AKN + 2∠BKL = 180°
⇒ ∠AKN + ∠BKL = 90°
⇒ ∠NKL = 90°
Similarly, we have
∠KLM = ∠LMN = ∠MNK = 90°
Hence, KLMN is a square.
Solution 26
Solution 27
Quadrilaterals Exercise Ex. 10A
Solution 1
Let the measure of the fourth angle = x°
For a quadrilateral, sum of four angles = 360°
⇒ x° + 75° + 90° + 75° = 360°
⇒ x° = 360° - 240°
⇒ x° = 120°
Hence, the measure of fourth angle is 120°.
Solution 2
Solution 3
Since AB || DC
Solution 4
Given:
Solution 5
Solution 6
Solution 7
Solution 8
Given: O is a point within a quadrilateral ABCD
Solution 9
Given: ABCD is a quadrilateral and AC is one of its diagonals.
Solution 10
Given: ABCD is a quadrilateral.
Quadrilaterals Exercise Ex. 10C
Solution 1
(i) In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
(ii) In ΔADC, R and S are the mid-points of sides CD and AD respectively.
From (i) and (ii), we have
PQ = SR and PQ ∥ SR
(iii) Thus, in quadrilateral PQRS, one pair of opposite sides are equal and parallel.
Hence, PQRS is a parallelogram.
Solution 2
Let ΔABC be an isosceles right triangle, right-angled at B.
⇒ AB = BC
Let PBSR be a square inscribed in ΔABC with common ∠B.
⇒ PB = BS = SR = RP
Now, AB - PB = BC - BS
⇒ AP = CS ….(i)
In ΔAPR and ΔCSR
AP = CS [From (i)
∠APR = ∠CSR (Each 90°)
PR = SR (sides of a square)
∴ ΔAPR ≅ ΔCSR (by SAS congruence criterion)
⇒ AR = CR (c.p.c.t.)
Thus, point R bisects the hypotenuse AC.
Solution 3
Solution 4
In ΔAOM and ΔCON
∠MAO = ∠OCN (Alternate angles)
AO = OC (Diagonals of a parallelogram bisect each other)
∠AOM = ∠CON (Vertically opposite angles)
∴ ΔAOM ≅ ΔCON (by ASA congruence criterion)
⇒ MO = NO (c.p.c.t.)
Thus, MN is bisected at point O.
Solution 5
Construction: Join diagonal QS. Let QS intersect MN at point O.
PQ ∥ SR and MN ∥ PQ
⇒ PQ ∥ MN ∥ SR
By converse of mid-point theorem a line drawn, through the mid-point of any side of a triangle and parallel to another side bisects the third side.
Now, in ΔSPQ
MO ∥ PQ and M is the mid-point of SP
So, this line will intersect QS at point O and O will be the mid-point of QS.
Also, MN ∥ SR
Thus, in ΔQRS, ON ∥ SR and O is the midpoint of line QS.
So, by using converse of mid-point theorem, N is the mid-point of QR.
Solution 6
PM is the bisector of ∠P.
⇒ ∠QPM = ∠SPM ….(i)
PQRS is a parallelogram.
∴ PQ ∥ SR and PM is the transversal.
⇒ ∠QPM = ∠MS (ii)(alternate angles)
From (i) and (ii),
∠SPM = ∠PMS ….(iii)
⇒ MS = PS = 9 cm (sides opposite to equal angles are equal)
Now, ∠RMT = ∠PMS (iv)(vertically opposite angles)
Also, PS ∥ QT and PT is the transversal.
∠RTM = ∠SPM
⇒ ∠RTM = ∠RMT
⇒ RT = RM (sides opposite to equal angles are equal)
RM = SR - MS = 12 - 9 = 3 cm
⇒ RT = 3 cm
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.
In ΔADC, S and R are the mid-points of sides AD and CD respectively.
In ΔABD, P and S are the mid-points of sides AB and AD respectively.
⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]
Thus, PQRS is a parallelogram.
Now, AC = BD (given)
⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]
Hence, PQRS is a rhombus.
Solution 17
In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
In ΔADC, R and S are the mid-points of sides CD and AD respectively.
From (i) and (ii),
PQ || RS and PQ = RS
Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.
So, PQRS is a parallelogram.
Let the diagonals AC and BD intersect at O.
Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.
Thus, in quadrilateral PMON, PM || NO and PN || MO.
⇒ PMON is a parallelogram.
⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)
⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)
⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)
⇒ ∠QPS = 90°
Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.
Hence, PQRS is a rectangle.
Solution 18
In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.
In ΔADC, S and R are the mid-points of sides AD and CD respectively.
In ΔABD, P and S are the mid-points of sides AB and AD respectively.
⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]
Thus, PQRS is a parallelogram.
Now, AC = BD (given)
⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]
Let the diagonals AC and BD intersect at O.
Now,
Thus, in quadrilateral PMON, PM || NO and PN || MO.
⇒ PMON is a parallelogram.
⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)
⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)
⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)
⇒ ∠QPS = 90°
Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.
Hence, PQRS is a square.